Solve for $x$ and $y$ using elimination. $\begin{align*}3x+7y &= 5 \\ -6x-2y &= -2\end{align*}$
We can eliminate $x$ when its corresponding coefficients are negative inverses. Recalling our knowledge of least common multiples, multiply the top equation by $2$ and the bottom equation by $1$ $\begin{align*}6x+14y &= 10\\ -6x-2y &= -2\end{align*}$ Add the top and bottom equations. $12y = 8$ Divide both sides by $12$ and reduce as necessary. $y = \dfrac{2}{3}$ Substitute $\dfrac{2}{3}$ for $y$ in the top equation. $3x+7( \dfrac{2}{3}) = 5$ $3x+\dfrac{14}{3} = 5$ $3x = \dfrac{1}{3}$ $x = \dfrac{1}{9}$ The solution is $\enspace x = \dfrac{1}{9}, \enspace y = \dfrac{2}{3}$.